3.1 Formula Mass and the Mole Concept

Formula mass and the mole concept are the foundation of chemical calculations. They let you quantify substances and connect the microscopic world of atoms to the macroscopic world you can actually weigh and measure.

These concepts are essential for stoichiometry, where you predict quantities in chemical reactions. Once you're comfortable with formula mass and moles, converting between grams, moles, and particles becomes second nature.

Formula Mass and Mole Concept

Formula Masses of Chemical Compounds

Formula mass is the total of all atomic masses for every atom in a chemical formula. It's expressed in atomic mass units (amu), and the number is numerically equal to the molar mass in grams per mole (g/mol).

To calculate formula mass, follow these steps:

1. Identify each element in the formula and its subscript (the subscript tells you how many atoms of that element are present).
2. Look up the atomic mass of each element on the periodic table.
3. Multiply each element's atomic mass by its subscript.
4. Add all the products together.

Example: Formula mass of glucose ($C_6H_{12}O_6$)

• Carbon (C): 6 × 12.01 amu = 72.06 amu
• Hydrogen (H): 12 × 1.01 amu = 12.12 amu
• Oxygen (O): 6 × 16.00 amu = 96.00 amu
• Formula mass = 72.06 + 12.12 + 96.00 = 180.18 amu (or 180.18 g/mol)

Example: Formula mass of sodium chloride ($NaCl$)

• Sodium (Na): 1 × 22.99 amu = 22.99 amu
• Chlorine (Cl): 1 × 35.45 amu = 35.45 amu
• Formula mass = 22.99 + 35.45 = 58.44 amu (or 58.44 g/mol)

Formula mass is also used to calculate percent composition, which tells you what fraction of a compound's mass comes from each element.

Mole Concept and Avogadro's Number

A mole is a counting unit for chemistry, just like "dozen" means 12. One mole of any substance contains exactly $6.022 \times 10^{23}$ particles. That number is called Avogadro's number, and the particles can be atoms, molecules, ions, or formula units depending on the substance.

The mole was originally defined as the number of atoms in exactly 12 grams of carbon-12. That's where the specific value of $6.022 \times 10^{23}$ comes from.

Molar mass is the mass of one mole of a substance, measured in g/mol. Numerically, it equals the formula mass. So if glucose has a formula mass of 180.18 amu, then one mole of glucose weighs 180.18 g.

Two key relationships to memorize:

• 1 mole = $6.022 \times 10^{23}$ particles
• 1 mole of a substance = its molar mass in grams

Example: One mole of water ($H_2O$) contains $6.022 \times 10^{23}$ water molecules and has a mass of 18.02 g.

Example: One mole of sodium ($Na$) contains $6.022 \times 10^{23}$ sodium atoms and has a mass of 22.99 g.

Conversions in Chemical Substances

All mole conversions rely on two conversion factors:

• 1 mole = molar mass in grams
• 1 mole = $6.022 \times 10^{23}$ particles

Mass ↔ Moles:

1. Mass to moles: divide mass by molar mass
2. Moles to mass: multiply moles by molar mass

Moles ↔ Particles:

1. Moles to particles: multiply moles by $6.022 \times 10^{23}$
2. Particles to moles: divide number of particles by $6.022 \times 10^{23}$

Mass ↔ Particles (two-step conversion):

1. Mass to particles: first convert mass to moles, then moles to particles
2. Particles to mass: first convert particles to moles, then moles to mass

Use dimensional analysis to set up these problems. Write out your units at every step and cancel them as you go. If your units don't cancel correctly, something's off.

Example: Convert 25.0 g of water to moles

The molar mass of $H_2O$ is 18.02 g/mol.

$25.0\,g\,H_2O \times \frac{1\,mol\,H_2O}{18.02\,g\,H_2O} = 1.39\,mol\,H_2O$

Example: How many oxygen atoms are in 0.500 mol of $O_2$?

This one is tricky because each $O_2$ molecule contains 2 oxygen atoms. You need to account for that:

$0.500\,mol\,O_2 \times \frac{6.022 \times 10^{23}\,molecules\,O_2}{1\,mol\,O_2} \times \frac{2\,atoms\,O}{1\,molecule\,O_2} = 6.02 \times 10^{23}\,atoms\,O$

Chemical Formulas and Stoichiometry

The empirical formula gives the simplest whole-number ratio of atoms in a compound. For example, glucose's empirical formula is $CH_2O$ (a 1:2:1 ratio of C, H, and O).

The molecular formula shows the actual number of each atom in one molecule. Glucose's molecular formula is $C_6H_{12}O_6$, which is exactly 6 times the empirical formula.

Stoichiometry uses mole ratios from balanced chemical equations to calculate how much reactant you need or how much product you'll get. The coefficients in a balanced equation tell you the mole ratios between substances.

The limiting reagent is the reactant that runs out first in a reaction. It determines the maximum amount of product that can form. The other reactant(s) are in excess.